Metamath Proof Explorer


Theorem dgrnznn

Description: A nonzero polynomial with a root has positive degree. (Contributed by Stefan O'Rear, 25-Nov-2014)

Ref Expression
Assertion dgrnznn ⊢ P ∈ Poly ⁡ S ∧ P ≠ 0 𝑝 ∧ A ∈ ℂ ∧ P ⁡ A = 0 → deg ⁡ P ∈ ℕ

Proof

Step Hyp Ref Expression
1 simpr ⊢ A ∈ ℂ ∧ P ⁡ A = 0 ∧ P = ℂ × P ⁡ 0 → P = ℂ × P ⁡ 0
2 1 fveq1d ⊢ A ∈ ℂ ∧ P ⁡ A = 0 ∧ P = ℂ × P ⁡ 0 → P ⁡ A = ℂ × P ⁡ 0 ⁡ A
3 simplr ⊢ A ∈ ℂ ∧ P ⁡ A = 0 ∧ P = ℂ × P ⁡ 0 → P ⁡ A = 0
4 fvex ⊢ P ⁡ 0 ∈ V
5 4 fvconst2 ⊢ A ∈ ℂ → ℂ × P ⁡ 0 ⁡ A = P ⁡ 0
6 5 ad2antrr ⊢ A ∈ ℂ ∧ P ⁡ A = 0 ∧ P = ℂ × P ⁡ 0 → ℂ × P ⁡ 0 ⁡ A = P ⁡ 0
7 2 3 6 3eqtr3rd ⊢ A ∈ ℂ ∧ P ⁡ A = 0 ∧ P = ℂ × P ⁡ 0 → P ⁡ 0 = 0
8 7 sneqd ⊢ A ∈ ℂ ∧ P ⁡ A = 0 ∧ P = ℂ × P ⁡ 0 → P ⁡ 0 = 0
9 8 xpeq2d ⊢ A ∈ ℂ ∧ P ⁡ A = 0 ∧ P = ℂ × P ⁡ 0 → ℂ × P ⁡ 0 = ℂ × 0
10 1 9 eqtrd ⊢ A ∈ ℂ ∧ P ⁡ A = 0 ∧ P = ℂ × P ⁡ 0 → P = ℂ × 0
11 df-0p ⊢ 0 𝑝 = ℂ × 0
12 10 11 eqtr4di ⊢ A ∈ ℂ ∧ P ⁡ A = 0 ∧ P = ℂ × P ⁡ 0 → P = 0 𝑝
13 12 ex ⊢ A ∈ ℂ ∧ P ⁡ A = 0 → P = ℂ × P ⁡ 0 → P = 0 𝑝
14 13 necon3ad ⊢ A ∈ ℂ ∧ P ⁡ A = 0 → P ≠ 0 𝑝 → ¬ P = ℂ × P ⁡ 0
15 14 impcom ⊢ P ≠ 0 𝑝 ∧ A ∈ ℂ ∧ P ⁡ A = 0 → ¬ P = ℂ × P ⁡ 0
16 15 adantll ⊢ P ∈ Poly ⁡ S ∧ P ≠ 0 𝑝 ∧ A ∈ ℂ ∧ P ⁡ A = 0 → ¬ P = ℂ × P ⁡ 0
17 0dgrb ⊢ P ∈ Poly ⁡ S → deg ⁡ P = 0 ↔ P = ℂ × P ⁡ 0
18 17 ad2antrr ⊢ P ∈ Poly ⁡ S ∧ P ≠ 0 𝑝 ∧ A ∈ ℂ ∧ P ⁡ A = 0 → deg ⁡ P = 0 ↔ P = ℂ × P ⁡ 0
19 16 18 mtbird ⊢ P ∈ Poly ⁡ S ∧ P ≠ 0 𝑝 ∧ A ∈ ℂ ∧ P ⁡ A = 0 → ¬ deg ⁡ P = 0
20 dgrcl ⊢ P ∈ Poly ⁡ S → deg ⁡ P ∈ ℕ 0
21 20 ad2antrr ⊢ P ∈ Poly ⁡ S ∧ P ≠ 0 𝑝 ∧ A ∈ ℂ ∧ P ⁡ A = 0 → deg ⁡ P ∈ ℕ 0
22 elnn0 ⊢ deg ⁡ P ∈ ℕ 0 ↔ deg ⁡ P ∈ ℕ ∨ deg ⁡ P = 0
23 21 22 sylib ⊢ P ∈ Poly ⁡ S ∧ P ≠ 0 𝑝 ∧ A ∈ ℂ ∧ P ⁡ A = 0 → deg ⁡ P ∈ ℕ ∨ deg ⁡ P = 0
24 orel2 ⊢ ¬ deg ⁡ P = 0 → deg ⁡ P ∈ ℕ ∨ deg ⁡ P = 0 → deg ⁡ P ∈ ℕ
25 19 23 24 sylc ⊢ P ∈ Poly ⁡ S ∧ P ≠ 0 𝑝 ∧ A ∈ ℂ ∧ P ⁡ A = 0 → deg ⁡ P ∈ ℕ